JEE Main 2024 (Online) 1st February Morning ShiftQuestion 1
Let $f:\mathbf{R}\to\mathbf{R}$ be defined as : $f(x)=\begin{cases}\frac{a-b\cos2x}{x^{2}}; & x<0\\ x^{2}+cx+2; & 0\le x\le1\\ 2x+1; & x>1\end{cases}$
If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differentiable then $m+a+b+c$ equals :
Differentiation and its Application
Solution
Concept. The left limit at $0$ exists only if $a=b$; continuity then fixes $a,b,c$; finally compare one-sided slopes. Formulas used. (1) $1-\cos2x=2\sin^2x$ and $\lim_{x\to0}\frac{\sin x}{x}=1$(2) $\frac{2\sin^2x}{x^2}=2-\frac23x^2+\cdots$Step 1 - At $x=0$. If $a\ne b$, $\frac{a-b\cos2x}{x^2}\to\pm\infty$. So $a=b$, and by (1) the left limit is $2a$. It must equal $f(0)=2$, so $a=b=1$. Step 2 - At $x=1$. $1+c+2=2+1$, so $c=0$. Step 3 - Corners. At $0$: by (2) the left slope is $0$; the right slope is $2x+c=0$. At $1$: left slope $2$, right slope $2$. So $m=0$. Step 4 - Value. $m+a+b+c=0+1+1+0=2$. Answer: (B)